Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $r = \dfrac{2y - 6}{y - 5} \times \dfrac{3y - 15}{y^2 - 10y + 21} $
First factor the quadratic. $r = \dfrac{2y - 6}{y - 5} \times \dfrac{3y - 15}{(y - 3)(y - 7)} $ Then factor out any other terms. $r = \dfrac{2(y - 3)}{y - 5} \times \dfrac{3(y - 5)}{(y - 3)(y - 7)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ 2(y - 3) \times 3(y - 5) } { (y - 5) \times (y - 3)(y - 7) } $ $r = \dfrac{ 6(y - 3)(y - 5)}{ (y - 5)(y - 3)(y - 7)} $ Notice that $(y - 5)$ and $(y - 3)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ 6\cancel{(y - 3)}(y - 5)}{ (y - 5)\cancel{(y - 3)}(y - 7)} $ We are dividing by $y - 3$ , so $y - 3 \neq 0$ Therefore, $y \neq 3$ $r = \dfrac{ 6\cancel{(y - 3)}\cancel{(y - 5)}}{ \cancel{(y - 5)}\cancel{(y - 3)}(y - 7)} $ We are dividing by $y - 5$ , so $y - 5 \neq 0$ Therefore, $y \neq 5$ $r = \dfrac{6}{y - 7} ; \space y \neq 3 ; \space y \neq 5 $